Problem 103 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The number of words that can be formed, with the letters of the work 'Pataliputra' without changing the relative order of the vowels and consonants, is (A) 3600 (B) 4200 (C) 3680 (D) None of these

Step-by-Step Solution

Answer

The number of words that can be formed is 3600, option (A).

Step 1: Identify the vowels and consonants

First, we need to separate the letters in the word 'Pataliputra' into vowels and consonants. The vowels are A, A, I, U, A, and the consonants are P, T, L, P, T, R.

Step 2: Count the number of vowels and consonants

Count the number of vowels and consonants identified in the word. We have 5 vowels (A, A, I, U, A) and 6 consonants (P, T, L, P, T, R).

Step 3: Calculate permutations for vowels

There are 5 vowels with A repeated 3 times. The number of ways to arrange these vowels is given by the formula for permutations of multiset: \[ \frac{5!}{3!} = \frac{120}{6} = 20 \]

Step 4: Calculate permutations for consonants

There are 6 consonants with P repeated 2 times and T repeated 2 times. The number of ways to arrange these consonants is: \[ \frac{6!}{2!2!} = \frac{720}{4} = 180 \]

Step 5: Calculate total permutations

The total number of words that can be formed without changing the relative order of vowels and consonants is the product of the permutations of vowels and consonants: \[ 20 \times 180 = 3600 \]

Step 6: Identify the correct option

The total number of words is 3600, which corresponds to option (A).

Key Concepts

Vowel Arrangement

When we talk about arranging vowels, we're essentially looking at how we can organize a set of specific letters. In the case of our exercise, the vowels in "Pataliputra" are A, A, I, U, A.
Here, the formula for arranging these vowels involves considering repeated elements. We use the **multiset permutation** formula, which is:

• Count the total number of vowels, which is 5 in this case.
• Identify the repetitions - A appears 3 times.

To arrange these vowels, the formula becomes: \[ \frac{5!}{3!} \] This simplifies to: \[ \frac{120}{6} = 20 \]
This means there are 20 distinct ways to order just the vowels from "Pataliputra."
Understanding this helps in grasping how repetitive elements affect permutations and is crucial for solving similar kinds of problems.

Consonant Arrangement

Arranging the consonants from the word "Pataliputra" also involves handling repeated elements. Here, the consonants are P, T, L, P, T, R.
Just like with vowels, we need to use the multiset permutation formula. The steps include:

• Count the total consonants, which are 6.
• Note repetitions: P and T each occur 2 times.

The formula used to calculate these permutations is:\[ \frac{6!}{2!2!} \] This simplifies to: \[ \frac{720}{4} = 180 \]
Thus, there are 180 unique ways to arrange the consonants. This part of the exercise illustrates how handling multiples of the same element can significantly reduce the total number of permutations.

Multiset Permutation

Multiset permutation is a concept used when there are repeated elements in a set. It modifies the factorial used in basic permutations to account for repetitions.
For a set with elements repeated multiple times, the formula is:\[ \frac{n!}{p_1! \times p_2! \times \ldots \times p_k!} \]

• \( n! \) is the factorial of the total number of items.
• Each \( p_i! \) represents the factorial of how many times a specific element is repeated.

For example, in our exercise:

• The vowels have a repeat element 'A' with 3 occurrences: \[ \frac{5!}{3!} \]
• The consonants have repeat elements 'P' and 'T', each with 2 occurrences: \[ \frac{6!}{2!2!} \]

This formula helps in simplifying the otherwise complicated task of calculating permutations where duplicates exist. This makes it a very valuable tool in combinatorics for students dealing with letters, numbers, or any group of items.